Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{r^3 + 10r^2 + 21r}{-2r^2 - 4r + 6} \div \dfrac{r^2 + 7r}{r + 1} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{r^3 + 10r^2 + 21r}{-2r^2 - 4r + 6} \times \dfrac{r + 1}{r^2 + 7r} $ First factor out any common factors. $k = \dfrac{r(r^2 + 10r + 21)}{-2(r^2 + 2r - 3)} \times \dfrac{r + 1}{r(r + 7)} $ Then factor the quadratic expressions. $k = \dfrac {r(r + 3)(r + 7)} {-2(r + 3)(r - 1)} \times \dfrac {r + 1} {r(r + 7)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac { r(r + 3)(r + 7) \times (r + 1)} { -2(r + 3)(r - 1) \times r(r + 7)} $ $k = \dfrac {r(r + 3)(r + 7)(r + 1)} {-2r(r + 3)(r - 1)(r + 7)} $ Notice that $(r + 3)$ and $(r + 7)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {r\cancel{(r + 3)}(r + 7)(r + 1)} {-2r\cancel{(r + 3)}(r - 1)(r + 7)} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $k = \dfrac {r\cancel{(r + 3)}\cancel{(r + 7)}(r + 1)} {-2r\cancel{(r + 3)}(r - 1)\cancel{(r + 7)}} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $k = \dfrac {r(r + 1)} {-2r(r - 1)} $ $ k = \dfrac{-(r + 1)}{2(r - 1)}; r \neq -3; r \neq -7 $